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OTOH, from the article: Sounds quite mystical to me.

That is being kind. This is Quantum Woo. Belongs in Religion rather than Physics, at best. Is Deepak Chopra involved? 🤪

The penetration depth of "cooking" microwaves into food is variable but typically about 3 cm. That for the IR from a grill is less than a millimetre. So, for an item less than a few cm thick, it is heated "in the middle" by the microwaves. The outside is heated more than the inside, but, unlike a conventional oven, heat is generated inside the item, rather than being conducted from the outside.

In some sense, yes. But it isn’t really a physical change that one would notice - spacetime remains smooth, regular, and locally Minkowskian everywhere (outside the singularity). What changes is mostly the physical meaning of the coordinates we use, relative to the exterior region. They now become spatial in nature, along the radial direction. Future means going “down” radially, past means going “up”. So the physical meaning of the r and t coordinates trade places. But again, it’s not something you would notice; spacetime looks just the same below as above the horizon. The only difference is its causal structure - below the horizon, all physically possible world lines (whether geodesics or not) terminate at the singularity. Locally, everything looks perfectly normal there, at least up to the point where tidal forces become noticeable. No, nothing special happens at the horizon at all - spacetime is perfectly regular there, and if you were to fall through it, you wouldn’t locally notice anything out of the ordinary. It’s really just a mathematical concept, not a physical entity. He can, because he’s in free fall. Visualise it like this (though it’s not really correct) - a photon emitted radially outwards very close to the horizon has a very slow radial (!) velocity wrt to the event horizon. On the other hand though, Pinocchio falls through the horizon at nearly the speed of light (wrt some outside reference), and thus meets the photon on the way. So it’s not like the photon necessarily propagates to his eyes, but rather that his eyes fall right to where the photon is. It’s kind of like jumping upwards in an elevator - you can put relative motion between yourself and the elevator floor, but both you and the elevator continue to move down regardless (maybe a stupid example, but you get my drift hopefully). Note that the situation is different if you’re not in free-fall. If Pinocchio, after the tip of his nose crosses the horizon, somehow fires magical thrusters that arrest his fall before he reaches the EH, then his nose will visually disappear for him (and get ripped away).

With the solution to the second problem in place, I decided to provide a solution to the generalised intermediate version of this problem, where the first and second problems are the two extremes. Because much of the solution to the second problem is unchanged in the solution to the generalised problem, I shall only include the parts of the solution that have changed. Let [math]\sigma[/math] be a parameter, [math]0 \leqslant \sigma \leqslant 1[/math], such that [math]\dfrac{\sigma \pi}{n}[/math] is the angle between the line joining the centre of the large circle to the centre of the outermost circle and the line joining the centre of the large circle to the centre of the nearest second outermost circles. Applying the law of cosines and continuing as shown in solution to the second problem: [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\sigma\pi}{n}) + 2 (1 + \cos\dfrac{\sigma\pi}{n}) = 0[/math] Note that the angle associated with [math]\cos^2\dfrac{\pi}{n}[/math] has a different origin to the angle associated with [math]1 + \cos\dfrac{\sigma\pi}{n}[/math]. [math]\phi = \dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\sigma\phi) + 2 (1 + \cos\sigma\phi) = 0[/math] [math]\phi \approx 0[/math] , [math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math] : [math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \sigma^2 \phi^2) + (4 - \sigma^2 \phi^2) = 0[/math] Solving the quadratic equation for [math](1 + \alpha)[/math] : [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(4 - \sigma^2 \phi^2)^2 - 4 (1 - \phi^2)(4 - \sigma^2 \phi^2)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 8 \sigma^2 \phi^2 + \sigma^4 \phi^4) - (16 - (16 + 4 \sigma^2) \phi^2 + 4 \sigma^2 \phi^4)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 4 \sigma^2) \phi^2 - (4 \sigma^2 - \sigma^4) \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = 2 - \sqrt{4 - \sigma^2}\>\phi[/math] [math]\alpha = 1 - \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math] [math]1 - \alpha^2 = 2 \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{4 - \sigma^2}}[/math] For problem #1, [math]\sigma = 0:[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{4}[/math] For problem #2, [math]\sigma = 1:[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]

It's a speculative thread, so I don't see why we can't still post sightings (as Moon was planning to do yesterday) and there could be debate as to their quality of data, what are reasonable testable hypotheses, etc. And I would like to see more academic institutions send (as happened in Texas with the university sending a team of science grad students and prof to look at the Marfa lights) investigation teams to study the anomalous and possibly extraordinary.

In systems theory, the measurement process is not a mystical matter. During a measurement, two systems come together. Described in: Quantum mechanical measurement in monistic systems theory. Doi:10.23756/sp.v11i2.1350